Chapter 1, Exercise 2.1: A homogeneous form of the Nullstellensatz

Prove the "homogeneous Nullstellensatz," which says that if $\mathfrak a\subseteq S$ is a homogeneous ideal, and if $f\in S$ is a homogeneous polynomial with $\deg f>0$, such that $f(P)=0$ for all $P\in Z(\mathfrak a)$ in $\mathbb P^n$, then $f^q\in\mathfrak a$ for some $q>0$. [Hint: Interpret the problem in terms of the affine $(n+1)$-space whose affine coordinate ring is $S$, and use the usual Nullstellensatz, (1.3A).]

The first case of this problem (a trivial case) is when $Z(\mathfrak a)=\emptyset$ in $\mathbb P^n$ and the condition that $f$ vanishes on $Z(\mathfrak a)$ is vacuous. In this case, in $\mathbb A^{n+1}$ either $Z(\mathfrak a)=\emptyset$ or $Z(\mathfrak a)=\{(0,\dots,0)\}$. In the first case $f$ vanishes on $Z(\mathfrak a)$ vacuously and thus $f^q\in\mathfrak a$ for some $q$. In the second case, since $f$ is homogeneous with $\deg f>0$ we have $f(0,\dots,0)=0$, hence the same holds and we are done.

The nontrivial case occurs when $Z(\mathfrak a)\neq\emptyset$ in $\mathbb P^n$. Interpreting this in $\mathbb A^{n+1}$, if $\mathfrak a$ is homogeneous then we can interpret $Z(\mathfrak a)$ as an affine cone: a union of the lines through the origin representing points through the origin. To prove this, let $P=(a_0,\dots,a_n)\in Z(\mathfrak a)$ be a point. The ideal $\mathfrak a$ has generators that are homogeneous, and thus it suffices to show that any homogeneous polynomial vanishing at $P$ vanishes on such a line. Indeed, since it is homogeneous it vanishes on any multiple $\lambda P$, and thus on the entire line. Hence each line is contained in $Z(\mathfrak a)$. The other inclusion is trivial.

Now, since $f$ vanishes on $Z(\mathfrak a)$ and $f$ is homogeneous, it also vanishes on any affine lines through the origin and thus on the cone $Z(\mathfrak a)\subseteq\mathbb A^{n+1}$. By the ordinary Nullstellensatz, $f^q\in\mathfrak a$ for some $q>0$.

Chapter 1, Exercise 1.12: Algebraic geometry breaks down over a non-algebraically closed field

Give an example of an irreducible polynomial $f\in\mathbb R[x,y]$, whose zero set $Z(f)$ in $\mathbb A^2_{\mathbb R}$ is not irreducible (cf. 1.4.2).

This is as easy and pathological as it sounds - take $f(x,y)=x^2+y^2+1$, then $Z(f)=\emptyset$ in $\mathbb A^2_{\mathbb R}$ which is vacuously not irreducible.

Chapter 1, Exercise *1.11: A variety that is not a local complete intersection

Let $Y\subseteq\mathbb A^3$ be the curve given parametrically by $x=t^3,y=t^4,z=t^5$. Show that $I(Y)$ is a prime ideal of height $2$ in $k[x,y,z]$ which cannot be generated by $2$ elements. We say $Y$ is not a local complete intersection-- cf. (Ex. 2.17).

This is a starred exercise, however we will provide a solution here since it is quite fun and yields some enlightening computations in commutative algebra. The difficulty in this problem is that it is not immediately obvious as to how we should parameterize this curve - indeed, the obvious method would no longer be based on a polynomial system and would use radicals which are disallowed in algebraic geometry. Thus we must turn to a more subtle way of dealing with this problem. Note that we have $\operatorname{height}\mathfrak p = \operatorname{codim}Y$ where $Y$ is the associated variety, and thus we need to show that $Y$ has dimension one.

To show $\dim Y=1$ (and that $I(Y)$ is prime in the process), we will construct a topological homeomorphism $\mathbb A^1\to Y$. Dimension and irreducibility are both preserved under homeomorphism and our desired conclusion would follow. The homeomorphism, as with most things in algebraic geometry, is canonical: by definition $Y$ is the image of $\varphi:\mathbb A^1\to\mathbb A^3$ taking $t\mapsto(t^3,t^4,t^5)$. To see that $\varphi$ is injective, suppose $\varphi(s)=\varphi(t)$ which simply means $$s^3=t^3\qquad s^4=t^4\qquad s^5=t^5.$$ If either variable is zero we are done (since we are working in an integral domain). If they are both nonzero, divide equation two by equation one to find $s=t$ again. Hence $\varphi$ is a bijection and it remains to show it is continuous. This means that if $W$ is any algebraic set in $\mathbb A^3$, then $\varphi^{-1}(W)$ is algebraic in $\mathbb A^1$. In other words, it is either finite or all of $\mathbb A^1$. Suppose that $\varphi^{-1}(W)$ is infinite, that is, there are infinitely many $t\in\mathbb A^1$ such that $(t^3,t^4,t^5)\in W$. If $W=Z(f_1,\dots,f_r)$ then each $f_i$ must vanish on infinitely many points, thus $f_i\equiv 0$ and $W=\mathbb A^3$. Consequently, $\varphi^{-1}(W)=\mathbb A^1$ which is indeed closed. Hence $\varphi$ is a homeomorphism and the first half of the exercise follows.

To speak about generators of $I(Y)$, we should give a concrete description of this ideal. Suppose $f\in A=k[x,y,z]$ such that $f(t^3,t^4,t^5)=0$ for all $t\in k$. In general we can write $$f(x,y,z)=\sum_{a_1,a_2,a_3}c_{a_1a_2a_3}x^{a_1}y^{a_2}z^{a_3}.$$ Plugging in $x=t^3,y=t^4,z=t^5$, we find $$f(t^3,t^4,t^5)=\sum_{a_1,a_2,a_3}c_{a_1,a_2,a_3}t^{3a_1+4a_2+5a_3}\stackrel{?}{=}0\quad\forall t\in k.$$ For the last condition to be true we must have that for any given value of $3a_1+4a_2+5a_3$ all valid $c_{a_1a_2a_3}$ sum to zero, in other words $$\tag{*}\sum_{3a_1+4a_2+5a_3=s}c_{a_1a_2a_3}=0$$ for any $s$. These polynomials make up $I(Y)$.

The next step is an extraordinarily tedious calculation, in which we work out the meaning of condition $(*)$ for $s\in\{0,1,2,\dots,10\}$. These eleven calculations are all extremely trivial: we will work out $s=5$ and $s=10$. For $s=5$, the only valid set of $a_i$ is $(0,0,1)$, and the sum collapses to a single term giving $c_{001}=0$. In other words, $f$ must have no term proportional to $z$. For $s=10$, the two valid combinations are $(2,1,0)$ and $(0,0,2)$ and thus $c_{210}+c_{002}=0$. In other words, the $x^2y$ and $z^2$ terms must have equal and opposite coefficients. In the end, we rule out the following terms: $$\text{constant},x,y,z,x^2,xy.$$ The following pairs must have equal and opposite coefficients: $$xz,y^2\qquad x^2,yz\qquad x^2y,z^2.$$ Thus, the polynomials $y^2-xz,x^2-yz,x^2y-z^2$ lie in $I(Y)$, and further generate it. They are minimal - if we were to remove one polynomial then the associated condition would no longer be assured. Thus $I(Y)$ cannot have fewer than three generators.

Chapter 1, Exercise 1.10: Topological properties of dimension

(a) If $Y$ is any subset of a topological space $X$, then $\dim Y\leq\dim X$.

(b) If $X$ is a topological space which is covered by a family of open sets $\{U_i\}$, then $\dim X=\sup\dim U_i$.

(c) Give an example of a topological space $X$ and a dense open subset $U$ with $\dim U<\dim X$.

(d) If $Y$ is a closed subset of an irreducible finite-dimensional topological space $X$, and if $\dim Y=\dim X$, then $Y=X$.

(e) Give an example of a noetherian topological space of infinite dimension.

(a) This is quite trivial: any closed irreducible chain in $Y$ is also such a chain in $X$, and as such cannot be longer than $\dim X$.

(b) By (a), each $\dim U_i$ is $\leq\dim X$ and thus $\sup\dim U_i\leq X$. Consider some chain $X_0\subset\dots\subset X_n$ in $X$, where $n=\dim X$. This has maximal length, thus $X_0$ must be a minimal closed irreducible subset of $X$. Thus $X_0$ is a point (more precisely, a singleton). We have $X_0\in U_i$ for some $i$. Then $X_j\cap U_i\neq\emptyset$ for each $j\leq n$, since each $X_j$ contains $X_0$, as does $U_i$. Further, each $U_i\cap X_j\subseteq X_j$ is irreducible and dense in $X_j$ by Exercise 1.6, and we see that $X_j\neq X_{j-1}$. Thus we can construct a chain $U_i\cap X_0\subset\dots\subset U_i\cap X_n$ of closed distinct irreducible sets in $U_i$, but this chain must have length $\leq\sup\dim U_i$.

(c) When looking for strange counterexamples involving "small" dense sets, the usual place to look is the Sierpinski space $X=\{0,1\}$ with open sets $\emptyset,\{1\},\{0,1\}$. Note that the longest applicable chain in $X$ is $\{1\}\subset\{0,1\}$, so $\dim X=2$, but $\dim\{1\}=1$ in $X$. Now, the only closed set containing $\{1\}$ is $X$ itself, so $\{1\}$ is dense but $\dim\{1\}<\dim\{0,1\}$.

(d) Suppose $Y\neq X$, and $Y_0\subset\dots\subset Y_n$ is a maximal chain in $Y$ (so $n=\dim Y$). Then $Y_0\subset\dots\subset Y_n\subset X$ is a longer chain in $X$ satisfying the hypotheses since $X$ is irreducible.

(e) Take any well-ordered set $S$ and give it the topology with closed sets being all initial segments. For example, choose $S=\mathbb N=\{0,1,2,\dots\}$. This satisfies a descending chain condition, since every chain will eventually reach $\{0\}$. However, it is infinite-dimensional since $\{0\}\subset\{0,1\}\subset\dots$ is a valid infinite ascending chain.

Chapter 1, Exercise 1.9: A lower bound on the dimension of irreducible components

Let $\mathfrak a\subseteq A=k[x_1,\dots,x_n]$ be an ideal which can be generated by $r$ elements. Then every irreducible component of $Z(\mathfrak a)$ has dimension $\geq n-r$.

This is an exercise that looks somewhat nontrivial, but ultimately reduces to a geometric realization of a lemma from commutative algebra generalizing Krull's Hauptidealsatz. For this reduction, we use the following consequence of Theorem 1.8Ab): if $Y$ is any affine variety in $\mathbb A^n$, then $$\dim Y = \dim A/I(Y) = n - \operatorname{height}I(Y).$$ In our case, if $Z(\mathfrak a)=\bigcup G_i$, then we need only prove that $\operatorname{height}I(G_i)\leq r$. In particular, the ideals $I(G_i)$ are minimal prime ideals over $\mathfrak a=I(Y)$ (where $Z(\mathfrak a)=Y$), so that the following lemma from commutative algebra (Krull's height lemma) will suffice.

Lemma 1. If $A$ is a noetherian ring, $\mathfrak a$ is a prime ideal generated by $r$ elements, and $\mathfrak p$ is a minimal prime ideal over $\mathfrak a$, then $\operatorname{height}\mathfrak p\leq r$.

This is true, and is sufficient.

Chapter 1, Exercise 1.8: Intersecting with a hypersurface decreases dimension by one

Let $Y$ be an affine variety of dimension $r$ in $\mathbb A^n$. Let $H$ be a hypersurface in $\mathbb A^n$, and assume that $Y\not\subseteq H$. Then every irreducible component of $Y\cap H$ has dimension $r-1$. (See (7.1) for a generalization.)

Note: It is implicit, but not stated, in this problem that $Y\cap H$ is nonempty. In the case $Y\cap H=\emptyset$ it is not even meaningful to talk about irreducible components since $\emptyset$ is not irreducible and as such has no decomposition.

This problem, probably the first truly nontrivial result given in an exercise thus far, works out most naturally algebraically because of the quotient decomposition of dimension. It is notable further for giving us an interesting application of Krull's Hauptidealsatz (Hartshorne's Theorem 1.11A).

First let us fix some notation: since $H$ is a hypersurface write $I(H)=(f)$, and let the irreducible components of $Y\cap H$ be denoted $G_i$.

Now, we will work in the affine coordinate ring $A(Y)=A/I(Y)$ of the variety $Y$. In particular, we claim that the canonical image of $f$ (the generator of $I(H)$) in this ring is neither a zero-divisor nor a unit. For the first statement, since $Y\not\subseteq H$ taking ideals gives $(f)\not\subseteq I(Y)$, so $f\notin I(Y)$. Thus, under the canonical map, $f$ is not sent to the additive identity of $A(Y)$. Now, we use the assertion that $Y\cap H\neq\emptyset$ to prove that $f$ is not a unit. If the image $\bar f$ was a unit in $A(Y)$ and we had a prime ideal $\mathfrak q$ containing $\bar f$, then $\mathfrak q=(1)$. Thus $\mathfrak q$ is not minimal - in particular, no minimal prime ideals in $A(Y)$ contain $\bar f$. We will show momentarily that this implies $Y\cap H=\emptyset$.

Our next goal is to study the algebraic representation of the $G_i$ under the correspondence. Clearly $I(G_i)$ is a prime ideal in $A$, but under the quotient map $A\to A(Y)$ the image $\mathfrak p_i$ is a prime ideal. Note that it contains $\bar f$, since $f\in I(G_i)$ and inclusion is preserved under quotient maps. It is minimal over $(\bar f)$ by the irreducibility of $G_i$, and in particular we see $\bar f$ is a unit. By the Hauptidealsatz, $\operatorname{height}\mathfrak p_i=1$ and by Theorem 1.8Ab) we see $$\dim A(Y)/\mathfrak p_i=\dim A(Y)-\operatorname{height}\mathfrak p_i=r-1.$$ Since $A(Y)$ is the coordinate ring of $G_i$ we see $\dim G_i=r-1$ as desired.

Chapter 1, Exercise 1.7: Basic properties of noetherian spaces

(a) Show that the following conditions are equivalent for a topological space $X$:
    (i) $X$ is noetherian; (ii) every nonempty family of closed subsets has a minimal element; (iii) $X$ satisfies the ascending chain condition for open subsets; (iv) every nonempty family of open subsets has a maximal element.

(b) A noetherian topological space is quasi-compact, i.e., every open cover has a finite subcover.

(c) Any subset of a noetherian topological space is noetherian in its induced topology.

(d) A noetherian space which is also Hausdorff must be a finite set with the discrete topology.

(a) We will verify that (i) is equivalent to (ii), that (iii) is equivalent to (iv), and that (i) is equivalent to (iii).

For the first equivalence, suppose $X$ is noetherian and let $\{X_\alpha\}$ be a family of closed sets that is nonempty. Pick a member $X_0\in\{X_\alpha\}$: if $X_0$ is minimal we are done. Otherwise, there is another set $X_1\in\{X_\alpha\}$ such that $X_1\subset X_0$. This process is iterated inductively. If no $X_i$ is minimal, then we construct a chain $\dots\subset X_2\subset X_1\subset X_0$ which does not stabilize, a contradiction. Conversely, if $X$ satisfies (ii) and $X_1\supset X_2\supset\dots$ is any descending chain, then the set $\{X_i\}$ has a minimal element and consequently the chain stabilizes.

The proof of the second equivalence is exactly the same, switching around the $\subset$ signs.

For the third equivalence, if $X$ is noetherian and $X_0\subset X_1\subset\dots$ is an ascending open chain in $X$, then $X_0^c\supset X_1^c\supset\dots$ is a descending closed chain which must stabilize, and since the complements stabilize so do the original sets. The other direction is identical, and the exercise is thus completed.

(b) Let $\{U_\alpha\}$ be any open cover for $X$ and consider the set $\Sigma$ of finite unions of the $U_\alpha$. Since $X$ is noetherian, $\Sigma$ has a maximal element $U=U_{\alpha_1}\cup\dots\cup U_{\alpha_r}$ by (a). For each $\alpha$, $U_\alpha\subseteq U$, and thus the elements of $U$ form a finite subcover.

(c) Equivalently by (a), we verify the ascending chain condition for open sets. Let $Y\subset X$ be arbitrary, and let $Y\cap U_0\subset U_1\subset\dots$ be an ascending chain of open sets in $Y$ where each $U_i$ is open in $X$. Then $U_0\subset U_1\subset\dots$ is an ascending chain in $X$ which stabilizes by assumption. Thus the intersection of this chain with $Y$ also stabilizes.

(d) Note that a noetherian space endowed with the discrete topology must be a finite set, since if $X$ was an infinite space with the discrete topology then we could trivially find non-stabilizing ascending chains of open subsets since every subset is open in $X$. Thus we need only show that a noetherian Hausdorff space has the discrete topology.

It is enough to show that each $Y\subset X$ is closed, or that $Y^c$ is open. For $y\in Y$, by Hausdorffness we can construct an open set $B_y$ such that the set of all $B_y$ as $y\in Y$ is an open cover for $Y$, and such that $B_y$ is disjoint from a neighborhood $C_x$ of any $x\in Y^c$. By (b) and (c), pick a finite subcover $\{B_{y_i}\}$ of $Y$. The $B_{y_i}$ correspond to subsets $C_{x_i}$ of $Y^c$. The set $C=\bigcup C_{x_i}$ is disjoint from $Y$, and contains all points of $Y^c$ by construction. Thus $Y^c=C$ is open.